Understanding Einstein Field Equations Through Tensor Calculus

[JDoolin]

Einstein Field Equations?

I have not been able to comprehend the Einstein Field Equation, the Stress Energy Tensor, the Ricci Tensor, the Einstein Tensor, and Christoffel Symbols. Though I am reasonably proficient at working with nested loops in programming, and I have a rudimentary knowledge of partial differential equations, there seems to be no source for spanning the gap between diff-eq, and Tensor calculus, and there also seems to be no source that carefully defines the units and terms of the Tensor calculus.

Am I correct in thinking that there is some analogy with the kinematic equations:

v=dx/dt
a=dv/dt​

Can one make similar equations with 4-displacements, 4-velocities, and 4-accelerations?


What I'd like to do is get from where I am now, to where I can sort of comprehend the articles on wikipeda on these topics.

Like for instance, I know the chain rule in partial differential equations

$$d f(x,y,z)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$​

Is this something that could be expressed in tensor notation?

 

[stevenb]

I have not been able to comprehend the Einstein Field Equation, the Stress Energy Tensor, the Ricci Tensor, the Einstein Tensor, and Christoffel Symbols. Though I am reasonably proficient at working with nested loops in programming, and I have a rudimentary knowledge of partial differential equations, there seems to be no source for spanning the gap between diff-eq, and Tensor calculus, and there also seems to be no source that carefully defines the units and terms of the Tensor calculus.

I think you can benefit from watching the Leonard Susskind Lectures on General Relativity, available on youtube. He is someone who has the knack for teaching effectively. Although, these lectures aren't the best source for the basic Tensor calculus information, they give the important intuitive explanations that will make it much easier to consult other more formal sources.

I also recommend the lectures by Alex Maloney (audio and notes available). Here you will get into more details and more formality to get everything rigourously correct. However, without the Susskind Lectures, this may be too big a jump for you, given the background you mentioned.

http://www.physics.mcgill.ca/~maloney/514/

 

[bcrowell]

there seems to be no source for spanning the gap between diff-eq, and Tensor calculus,

Have you had vector calculus?

and there also seems to be no source that carefully defines the units and terms of the Tensor calculus.

http://www.lightandmatter.com/genrel/" is my own shot at introducing tensors from scratch. You might also find it helpful to read Exploring Black Holes, by Taylor and Wheeler. It discusses a lot of GR *without* tensors, and if you can understand all of that physics, you should have a very good foundation for understanding the same material using tensors.

Am I correct in thinking that there is some analogy with the kinematic equations:

v=dx/dt
a=dv/dt​

Can one make similar equations with 4-displacements, 4-velocities, and 4-accelerations?

Yes, these have direct analogies in terms of 4-vectors: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.2 But there is no real analogy with the Einstein field equations. (Don't know if that's what you were asking.)

To understand the physical content of the field equations, try the Feynman lectures and/or Penrose's The Road to Reality.

Like for instance, I know the chain rule in partial differential equations

$$d f(x,y,z)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$​

Is this something that could be expressed in tensor notation?

Yes. For an example, see http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.7 , subsection 5.7.2.

 

[Fredrik]

I have not been able to comprehend the Einstein Field Equation, the Stress Energy Tensor, the Ricci Tensor, the Einstein Tensor, and Christoffel Symbols. Though I am reasonably proficient at working with nested loops in programming, and I have a rudimentary knowledge of partial differential equations, there seems to be no source for spanning the gap between diff-eq, and Tensor calculus, and there also seems to be no source that carefully defines the units and terms of the Tensor calculus.

If you want to really know this stuff, you will at some point have to study differential geometry. I like Lee's books. "Introduction to smooth manifolds" covers the basics (and a lot more). The stuff about connections, covariant derivatives, parallel transport and curvature is in "Riemannian manifolds: an introduction to curvature".

Am I correct in thinking that there is some analogy with the kinematic equations:

v=dx/dt
a=dv/dt​

Can one make similar equations with 4-displacements, 4-velocities, and 4-accelerations?

If C:[a,b]→M is a smooth curve, then its tangent vector at C(t) is denoted by $$\dot C(t)$$ and defined by

$$\dot C(t)f=(f\circ C)'(t)$$

for all smooth functions f from an open neighborhood of C(t) into the real numbers. So the tangent vector of a curve, at a specific point on the curve, is an operator that takes functions to numbers. When C is the mathematical representation of the motion of a particle in a 4-dimensional spacetime, its tangent vector field $$\dot C$$ is called its 4-velocity. (In special relativity, these definitions can be simplified).

Like for instance, I know the chain rule in partial differential equations

$$d f(x,y,z)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz$$​

Is this something that could be expressed in tensor notation?

There's this calculation:

$$df|_p (v)=v(f)=v(x^i)\frac{\partial}{\partial x^i}\bigg|_p f =dx^i|_p(v)\frac{\partial}{\partial x^i}\bigg|_p f$$

(where v is an arbitrary tangent vector at p), which implies

$$df=\frac{\partial f}{\partial x^i} dx^i$$

but to really explain this requires a lot more time than I'm willing to spend tonight. Some of my earlier posts may help, e.g. this one and this one. Note that $$\{dx^i|_p\}$$ is a basis for the cotangent space at p. Specifically, it's the dual basis of $$\bigg\{\frac{\partial}{\partial x^i}\bigg|_p\bigg\}$$

 

[JDoolin]

Hey, I just wanted to say, there's a lot of really good stuff here. It will keep me busy for a while. Thanks to Fredrik, bcrowell, and stevenb

 

[JDoolin]

I think you can benefit from watching the Leonard Susskind Lectures on General Relativity, available on youtube. He is someone who has the knack for teaching effectively. Although, these lectures aren't the best source for the basic Tensor calculus information, they give the important intuitive explanations that will make it much easier to consult other more formal sources.

I also recommend the lectures by Alex Maloney (audio and notes available). Here you will get into more details and more formality to get everything rigourously correct. However, without the Susskind Lectures, this may be too big a jump for you, given the background you mentioned.

http://www.physics.mcgill.ca/~maloney/514/

The Leonard Susskind lectures have been really helpful.

Let's see if I've got any of this right:

The covariant tensors (indices downstairs) are like gradients; vector quantities that are functions of position.

The contravariant tensors (indices upstairs) are like differential elements; quantities that represent the positions themselves.

 

[stevenb]

The Leonard Susskind lectures have been really helpful.

Let's see if I've got any of this right:

The covariant tensors (indices downstairs) are like gradients; vector quantities that are functions of position.

The contravariant tensors (indices upstairs) are like differential elements; quantities that represent the positions themselves.

Sounds like you're off to a good start.

Just a word of caution about the lectures. They are very streamlined to give you the most critical concepts so that you have a sense of the landscape. From there, it will be much easier to get into the nitty gritty details. As an example, if I remember correctly, he presents the formula for covariant derivative of a covariant tensor with a sign error. There are some other minor mistakes too. Keep in mind that even the best physicists gets signs and factors of two wrong, ... perhaps even more often than the average Joe, for whatever reason.

Consider this lecture series like that "thin book" on a library shelf. One of my favorite professors back in school had a useful piece of advice for his students. He gave it in the form of a question by saying, "If you go to the library to get a book to learn a new subject, which book should you take"? Students would then throw out a bunch of answers until someone finally said, "Pick the thickest book!", to which he would reply, "No, pick the thinnest book."

His point was simply that it's better to start with a brief overview of a subject, hitting the major concepts, before getting into the details. Anyway, the Susskind lectures (even the ones on other topics), are those "thin books" you can read as introductions to new topics. Of course, the book needs to be good as well as thin, but Susskind succeeds there as well, in my opinion.

 

[JDoolin]

I haven't had a chance to look at this very much in the last couple of weeks, but maybe something that I was teaching in algebra might apply to rank 0 tensors.

I want to take a function f(x) and stretch it along the y-axis by a factor of V, stretch it along the x-axis by a factor of H, then transpose it along the x and y axis, by x₀ and y₀.

The end product looks like

$$\[g(x)=V f(\frac {1}{H}(x-x_0))+y_0\]$$​

Are there contravariant and covariant 0 degree tensors involved here?

 

[Mentz114]

Are there contravariant and covariant 0 degree tensors involved here?

What follows may be dumbing down the subject, but it helped me when I was first learning about curved spaces and tensor calculus.

Imagine two axes, representing say, x and y directions. One usually thinks of two straight lines at right angles crossing at the origin x=0, y=0. In this case defining a position vector from the origin to a point x₀, y₀ is trivial i.e. drop a perpendicular from the point to each axis and read off the value. Now imagine that the axes are not at right angles, and curve smoothly. One can still drop a perpendicular to the axis and read off a value. But there's a nother set of values available - by drawing a line through the point parallel to each axis and reading off the value where this line cuts the other axis.
The two sets of values obtained thus are called the perpendicular projected components, and the parallel projected components.

The motivation for this is that the quantity defined as the inner product of these two 'vectors' is a geometric invariant. It will not change under coordinate transformations.

Writing the perpendicular vector as X_a, and the parallel as X^a, the quantity X_aX^a=X₁X¹+X₂X² = L² is invariant.

This means that if we can write our physical observables in curved space as scalars formed by contracting tensors, they will be covariant under general coordinate transformation.

To summarize : to define an invariant length in curved space, we must allow two ways of expressing positions, the covariant and contravariant vector.

And, the metric of the space, a rank-2 tensor can be used to transform covariant -> contravariant and vice-versa, viz. X^a=g^abX_b

 

[JDoolin]

What follows may be dumbing down the subject, but it helped me when I was first learning about curved spaces and tensor calculus.

Imagine two axes, representing say, x and y directions. One usually thinks of two straight lines at right angles crossing at the origin x=0, y=0. In this case defining a position vector from the origin to a point x₀, y₀ is trivial i.e. drop a perpendicular from the point to each axis and read off the value. Now imagine that the axes are not at right angles, and curve smoothly. One can still drop a perpendicular to the axis and read off a value. But there's a nother set of values available - by drawing a line through the point parallel to each axis and reading off the value where this line cuts the other axis.
The two sets of values obtained thus are called the perpendicular projected components, and the parallel projected components.

The motivation for this is that the quantity defined as the inner product of these two 'vectors' is a geometric invariant. It will not change under coordinate transformations.

Writing the perpendicular vector as X_a, and the parallel as X^a, the quantity X_aX^a=X₁X¹+X₂X² = L² is invariant.

This means that if we can write our physical observables in curved space as scalars formed by contracting tensors, they will be covariant under general coordinate transformation.

To summarize : to define an invariant length in curved space, we must allow two ways of expressing positions, the covariant and contravariant vector.

And, the metric of the space, a rank-2 tensor can be used to transform covariant -> contravariant and vice-versa, viz. X^a=g^abX_b

Okay. Let's see if I can envision an example. If we've got the idea we can curve space, we could turn an arc-lengths of several concentric circles into a parallel straight lines, and we could take the radial lines rom the center, and also make parallel straight lines from them.

Or, we could take hyperbolas and lines from the origin (see attached) and convert them to parallel straight lines. If we can come up with precise mathematical transformations that do these things, then we could discuss which elements are covariant, contravariant, and invariant, based on how the transformation is accomplished, right?

 

[atyy]

Yes, it's basically partial differential equations. Like Maxwell's equations written with the 4-vector potential - the vector potential is not observable, and there are many vector potentials corresponding to the same physical situation or E and B fields. So in addition to the differential equations, there is a rule saying which potentials correspond to the same physical situation. This identification of different solutions as the same is what we call geometry - the true object that remains unchanged by different descriptions of it. In practical calculations, one can gauge fix.

 

[Mentz114]

Okay. Let's see if I can envision an example. If we've got the idea we can curve space, we could turn an arc-lengths of several concentric circles into a parallel straight lines, and we could take the radial lines rom the center, and also make parallel straight lines from them.

Or, we could take hyperbolas and lines from the origin (see attached) and convert them to parallel straight lines. If we can come up with precise mathematical transformations that do these things, then we could discuss which elements are covariant, contravariant, and invariant, based on how the transformation is accomplished, right?

Sorry, I can't see the relevance of the above to what I wrote. A (crude) diagram is attached to illustrate what I'm saying. The key is that the axes are curved. Notice that if the curved axes deform back to the usual Euclidean axes, the parallel (contravariant) and perpendicular ( covariant) components of the vector coincide. So in flat space you can put tensor indexes high or low.

 

[JDoolin]

It's the concepts of invariant, covariant, contravariant that throw me. From Special Relativity I know that distances, times, simultaneity are all observer-dependent quantities.

But proper time and proper distance between events are invariant. However, when you look at the proper time between events in a gravitational field, this is no longer invariant. If you are looking down,you see things moving in slow-motion. Looking up, you see things moving in fast motion.

If you told me we were working on a problem to find the work, and said $$W= \vec{F}\cdot \vec{d}$$.

I will take a guess and say Work is invariant, Force is covariant, and distance is contravariant. Is that right?

What are some examples of invariant quantities, and do these remain invariant in gravitational fields?

 

[Mentz114]

If you told me we were working on a problem to find the work, and said $W= \vec{F}\cdot \vec{d}$.

I will take a guess and say Work is invariant, Force is covariant, and distance is contravariant. Is that right?

What are some examples of invariant quantities, and do these remain invariant in gravitational fields?

If F and d were contravariant and covariant then W would be invariant under coordinate transformations.

The points you raise are about physics in curved space-time, which is not GR. GR is the theory of how matter curves space-time, and it turns out that pseudo-Riemannian geometry is the correct way to formulate it, and the field equations are the 'final' result.

I realize that just seeing how covariant and contravariant vectors arise in curved spacetime is not enough to understand tensors. I may be overplaying the invariance card. You will have to think abstractly to understand the Riemannian manifold, which is about worldlines and the vectors and vector fields associated with them. The kernel of thing is 'curvature', and you've been looking at that in earlier posts.

Two points about physics in curved spacetime. One obvious adjustment is that the distance² between points (a,b) and (c,d) is no longer (a-c)^2 + (b-d)^2 ( i.e. dx^2+dy^2) but is now given by integrating ds²=Sum g_abdx^adx^b. The most important is that the usual derivative is replaced by the covariant derivative.

A good book on GR is Stephani's "General Relativity : An Introduction to the Theory of the Gravitational Field" (1986) . It really does start easy with 'Force-free motion ... in Newtonian mechanics'.

[edit] cut manifest falsehoods.

 

[JDoolin]

Let me ask some basic questions that I'm getting out of Misner/Thorne/Wheeler; MTW.

If you watch an ant walking around an apple, do you really believe that ant thinks its walking in a straight line? Do you think the apple is flat and geometry is somehow curved, or do you think the apple is actually meaningfully round in some global reference frame?

If you see light bending around the sun in a solar eclipse, do you think that the light traveled in a straight line, but space was somehow bent, or do you think that the light actually meaningfully changed directions in a global refence frame?

If you see two satellites travel along geodesic paths and somehow meet at the same point twice, do you think they both traveled straight lines in some warped space-time, or do you think that the satellites actually meaningfully changed directions in a global reference frame?

From the writing, I have this sense that the authors MTW really truly believe there is no global reference frame, and that in each instance, the ant, the light, and the satellites really are moving in straight lines. They manage to convince themselves of this by saying that the light and the satellite feel no forces; thus they must be traveling in straight lines.

These are the arguments they are using to entice me down the rabbit hole, as Einstein said "Why were another seven years required for the construction of the general theory of relativity? The main reason lies in the fact that it is not so easy to free oneself from the idea that coordinates must have an immediate metrical meaning."

I have no such desire to free myself from this idea. To me, any legitimate coordinate system should have immediate metrical meaning. And all coordinate transformations simply change the metrical meanings.

Two points about physics in curved spacetime. One obvious adjustment is that the distance² between points (a,b) and (c,d) is no longer (a-c)^2 + (b-d)^2 ( i.e. dx^2+dy^2) but is now given by integrating ds²=Sum g_abdx^adx^b. The most important is that the usual derivative is replaced by the covariant derivative.

When I think of the distance between two points, I think of the shortest distance between two points. Integrating along the path to find the distance between two points represents the distance traveled along the path between the two points. I don't see how this is a change in space-time. It is a change in what you mean by distance.

 

[Fredrik]

If you watch an ant walking around an apple, do you really believe that ant thinks its walking in a straight line? Do you think the apple is flat and geometry is somehow curved, or do you think the apple is actually meaningfully round in some global reference frame?

Don't take analogies too seriously. You have to imagine a two-dimensional ant and a mathematical apple for this to be really accurate.

If you see light bending around the sun in a solar eclipse, do you think that the light traveled in a straight line, but space was somehow bent, or do you think that the light actually meaningfully changed directions in a global refence frame?

It's a "straight line" (technically a geodesic) in spacetime, not space. You need a coordinate system just to define which slice of spacetime to call space, and in the coordinate systems that would be convenient to use in this situation, the path through space is curved.

If you see two satellites travel along geodesic paths and somehow meet at the same point twice, do you think they both traveled straight lines in some warped space-time, or do you think that the satellites actually meaningfully changed directions in a global reference frame?

It sounds like you're asking us to ignore what the theory is saying and tell you what we think actually happens. Since I don't think my intuition is more accurate than general relativity, I can't even tell you what it would mean to do that.

To me, any legitimate coordinate system should have immediate metrical meaning. And all coordinate transformations simply change the metrical meanings.

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

When I think of the distance between two points, I think of the shortest distance between two points. Integrating along the path to find the distance between two points represents the distance traveled along the path between the two points.

Yes, and the result is the length of the curve. "Length" is the appropriate word here, not "distance".

I don't see how this is a change in space-time. It is a change in what you mean by distance.

In spacetime, the function we're integrating involves a square root of something that can be positive or negative. This means that we can define the "proper length" of a curve only for curves such that the quantity under the square root is positive everywhere on the curve. But we can also define a similar quantity, "proper time", for curves such that the quantity under the square root is negative everywhere on the curve. We just flip the sign of that quantity in the definition, so that we have something positive under the square root.

 

[JDoolin]

Don't take analogies too seriously. You have to imagine a two-dimensional ant and a mathematical apple for this to be really accurate.


It's a "straight line" (technically a geodesic) in spacetime, not space. You need a coordinate system just to define which slice of spacetime to call space, and in the coordinate systems that would be convenient to use in this situation, the path through space is curved.


It sounds like you're asking us to ignore what the theory is saying and tell you what we think actually happens. Since I don't think my intuition is more accurate than general relativity, I can't even tell you what it would mean to do that.


A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

With some requirements, I hope. You can't just take arbitrary sets of 4 numbers and assign them randomly to the events and call it a coordinate system. The coordinates have to be somehow meaningful.

If you can establish a one-to-one and onto relationship between the coordinate systems then you have a global coordinate system that can be viewed many different and valid ways.

Yes, and the result is the length of the curve. "Length" is the appropriate word here, not "distance".


In spacetime, the function we're integrating involves a square root of something that can be positive or negative. This means that we can define the "proper length" of a curve only for curves such that the quantity under the square root is positive everywhere on the curve. But we can also define a similar quantity, "proper time", for curves such that the quantity under the square root is negative everywhere on the curve. We just flip the sign of that quantity in the definition, so that we have something positive under the square root.

But how do you determine an appropriate way to find the proper length, using an integral of a curve? If you have spacelike separation, there is no valid "path" between the events. In any case, the coordinates of the events are well defined in any given coordinate system. You don't need to do an integral to determine the distance or the time between them.

The space-time interval between events tells you properties only of a hypothethical object, that would have certain properties if it were there. The vast majority of event pairs don't have an object in between them. And the events themselves exist and have locations and times in a global reference frame, describable without doing any integration at all.

 

[Mentz114]

But how do you determine an appropriate way to find the proper length, using an integral of a curve? If you have spacelike separation, there is no valid "path" between the events. In any case, the coordinates of the events are well defined in any given coordinate system. You don't need to do an integral to determine the distance or the time between them.

If you have a curve y=f(x), the the length along the curve between two points is (as you know)

$$s=\int_{p_0}^{p_1} \sqrt{1+(dy/dx)^2} dx$$

or something similar.

We have to think in terms of curves because worldlines are curves. In Lorentzian space ( t,x) we define the proper length as

$$s=\int_{p_0}^{p_1} \sqrt{1-(dy/dt)^2} dt$$

Most of your remarks are based on flat-space. If the 'axes' of a space are not Euclidean straight lines then there is no canonical distance and you need to define lengths of curve sections.

 

[JDoolin]

If you have a curve y=f(x), the the length along the curve between two points is (as you know)

$$s=\int_{p_0}^{p_1} \sqrt{1+(dy/dx)^2} dx$$

or something similar.

We have to think in terms of curves because worldlines are curves. In Lorentzian space ( t,x) we define the proper length as

$$s=\int_{p_0}^{p_1} \sqrt{1-(dy/dt)^2} dt$$

Most of your remarks are based on flat-space. If the 'axes' of a space are not Euclidean straight lines then there is no canonical distance and you need to define lengths of curve sections.

In your equation you've got events p0, and p1. What if you don't need s? What if you're more interested in what is happening in your own frame of reference, and you're not really concerned about some nonexistent ruler or particle traveling between these events?

For instance, what if p0 and p1 are two supernova explosions, that happen in two different parts of the sky several thousand years apart. Certainly we could calculate the space-time interval between those two events, but what possible use would it have?

It would be an answer to this question: (What is the distance between those events in a reference fram where the two events appear to be simultaneous?) but that question is completely irrelevant unless you happen to be traveling in just the right velocity where the two events were simultaneous.

With the Euclidian distance [tex]r^2=x^2+y^2+z^2[/itex], there are all kinds of things you can do with it: Work = Force dot distance, Torque =Force Cross Distance, velocity = distance over time, etc, etc.

But with this s distance $s^2=t^2-x^2-y^2-z^2$, what good physics comes from this? I think none.

I think that t, x, y, z are contravariants. s is invariant. But if you want to talk about velocity, momentum, these are covariant quantities, based on properties of matter within the space. Perhaps it is here where general relativity can shine. But it is not because somehow you warped the space; up down, left, right, forward, backward, future and past, are still defined in a global inertial reference frame. You KNOW you're spinning, so what was once left is now forward. But that's a global change! EVERYTHING that was once to your left is now forward. You can't make the claim that the universe is somehow only "locally Euclidian," when everyday experience tells you that turning your head allows you to see what's over to your left or right--no matter how far away it is. Extreme distances do not free events from the laws of Euclidian rotation. Why should extreme distance free events from the laws of Lorentz Transformation?

Yet that's what the argument seems to be in Chapter 1 of Misner Thorne Wheeler. Well, technically, they argue that "the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian" and "the geometry of a tiny thumbprint on the apple is Euclidian." (I may be making thehttp://www.jimloy.com/logic/converse.htm" ). However, with Chapter 7 entitled "Incompatibilty of Gravity and Special Relativity" I suspect they also meant the inverse: that Lorentzian spacetime can only exist in sufficiently limited regions.

 

[JDoolin]

MTW Chapter 7 Section 3:

"... Schhild's [argument uses] a global Lorentz frame tied to the Earth's center. It makes no demand that free particles initially at rest remain at rest in this global Lorentz frame"

Now I see at least part of the problem.

You can't tie a global inertial frame to anything unless it never accelerates.

The title of the chapter is "Incompatibility of Gravity and Special Relativity" and it is "Track-2" material, meaning that except for the obvious mistake of trying to tie a global inertial reference frame to an accelerating object, it mostly uses notation that I don't know.

 

[Mentz114]

But with this s distance $s^2=t^2-x^2-y^2-z^2$ , what good physics comes from this? I think none.

I don't think you can mean this.

For instance, what if p0 and p1 are two supernova explosions, that happen in two different parts of the sky several thousand years apart. Certainly we could calculate the space-time interval between those two events, but what possible use would it have?

Probably none, but only because you've chosen a bad example.

It would be an answer to this question: (What is the distance between those events in a reference frame where the two events appear to be simultaneous?) but that question is completely irrelevant unless you happen to be traveling in just the right velocity where the two events were simultaneous.

How would you calculate how long it takes light to get from the source to an observer ? How would you calculate what time is showing on the observers clocks ?
To answer these questions you need to know the worldline of the source, and of the observers, and the metric of the spacetime. The local clock times are integrals of the proper length ( from which no good physics can arise !) of the observers worldlines, and getting the light travel times mean working with null geodesics. Of course, if a galaxy deflects the light things get even more interesting.

Because the coordinate speed of light can vary in curved spacetimes, SR is no longer applicable globally. What we have to work with is local frames which are transported along worldlines. We can define these as inertial frames if they are geodesics, provided we don't make the volume they cover too large ( I see you are getting this from MTW).

The key to all this is intrinsic curvature. In another thread you started analysing curvature but it's not as straightforward as you think, judging by what you've written. It is not difficult to define the curvature of a line. For a 2D surface embedded in 3D it can also be done, but you obviously need 2 numbers at least to describe the 2D curvature. To define curvature in 4D we have to consider 3D hyperslices, each of which can be sliced into 3 orthogonal embedded 2D sufaces - my point being that it takes a whole bunch of numbers to fully describe intrinsic curvature. Up to 20 in fact. In order for curvature to be covariant it must be expressed as tensors.

Keep reading. Although what you've quoted from MTW sounds to me like a waffly attempt to justify what needs no justification. But I've never like fat books that try to explain everything. Good luck with it, it will probably get to the point eventually.

 

[Fredrik]

With some requirements, I hope. You can't just take arbitrary sets of 4 numbers and assign them randomly to the events and call it a coordinate system. The coordinates have to be somehow meaningful.

It has to be a continuous bijection from an open subset of the manifold into an open subset of $\mathbb R^n$, with a continuous inverse. I don't know how you would define "meaningful", but I wouldn't say that coordinates have to be meaningful.

Two coordinate systems $x:U\rightarrow\mathbb R^n$ and $y:V\rightarrow\mathbb R^n$ are said to be smoothly compatible if $U\cap V=\emptyset$ or $x\circ y^{-1}$ is smooth. (That means that its partial derivatives exist up to arbitrary order). A set of smoothly compatible coordinate systems is called an atlas. An atlas is said to be maximal if it's not a proper subset of another atlas. The set of coordinate systems on a smooth manifold is required to be a maximal atlas.

There are no other requirements than that (unless of course I forgot some minor technicality).

If you can establish a one-to-one and onto relationship between the coordinate systems then you have a global coordinate system that can be viewed many different and valid ways.

Many manifolds don't have a global coordinate system. The simplest example is a sphere.

But how do you determine an appropriate way to find the proper length, using an integral of a curve?

I don't understand the question. The proper length of a curve is defined as an integral.

If you have spacelike separation, there is no valid "path" between the events.

Sure there is.

In any case, the coordinates of the events are well defined in any given coordinate system. You don't need to do an integral to determine the distance or the time between them.

Right, but the coordinate distance and the coordinate time difference between the events depend on the coordinate system. Proper length and proper time depend only on the curve.

 

[JDoolin]

Susskind Lecture 5​

First ten minutes

Struggling my way through this topic.

We want to discuss the properties of geometries of various kinds. We can talk about properties of the Earth without introducing coordinates. But it is convenient to be able to talk about components of vectors, tensors and so forth are the way to describe properties, in general. We want relationships which are independent of the coordinates.

The fact that the Earth is almost a sphere is independent of coordinates
The fact that Mt. Everest is a lump on the Earth is idependent of coordinates
Properties are coordinate independent. But what we want to do is express those facts in such a way that the numerical left side and right side do depend on the coordinates, but the fact that they are equal does not depend on the coordinates.

If we can write our laws of physics in such a way that when properties are equal in one coordinate system then they are necessarily equal in all coordinate systems, then the physics does not depend on the coordinate system.

$$W_{nm}(x)=V_{nm}(x) \Leftrightarrow W_{nm}(y)=V_{nm}(y)$$

For instance, we could have
$$V_{nm}(x)= \left( \begin{matrix} V_{xx}(x,y,z) & V_{xy}(x,y,z) & V_{xz}(x,y,z) \\ V_{yx}(x,y,z) & V_{yy}(x,y,z) & V_{yz}(x,y,z) \\ V_{zx}(x,y,z) & V_{zy}(x,y,z) & V_{zz}(x,y,z) \end{matrix} \right)$$and $$V_{nm}(y) = \begin{pmatrix} V_{rr}(r,\theta,\varphi) & V_{r\theta}(r,\theta,\varphi) & V_{r\varphi}(r,\theta,\varphi) \\ V_{\theta r}(r,\theta,\varphi) & V_{\theta \theta}(r,\theta,\varphi) & V_{\theta \varphi}(r,\theta,\varphi) \\ V_{\varphi r}(r,\theta,\varphi) & V_{\varphi \theta}(r,\theta,\varphi) & V_{\varphi \varphi}(r,\theta,\varphi) \end{pmatrix}$$

Obviously $V_{nm}(x) \neq V_{nm}(y)$ but the fact that V=W will be true in both coordinate systems.

The objects that have this property are called tensors.

We'll simplify this question a little bit, just considering a vector instead of a 3X3 tensor.
$$V_n(y) = \frac{\partial x^r}{\partial y^n}V_r(x)$$

Expanded out, this equation looks something like this:

$$\begin{pmatrix} V_r(r,\theta,\varphi) \\ V_\theta(r,\theta,\varphi) \\ V_\varphi(r,\theta,\varphi) \\ \end{pmatrix} =\begin{pmatrix} \frac{\partial x}{\partial r} V_x + \frac{\partial y}{\partial r} V_y+\frac{\partial z}{\partial r} V_z \\ \frac{\partial x}{\partial \theta} V_x + \frac{\partial y}{\partial \theta} V_y+\frac{\partial z}{\partial \theta} V_z \\ \frac{\partial x}{\partial \varphi} V_x + \frac{\partial y}{\partial \varphi} V_y+\frac{\partial z}{\partial \varphi} V_z \end{pmatrix}$$


Another way of saying that $V_{nm}(x)= W_{nm}(x)$ is to say that $V_{nm}(x) - W_{nm}(x)=0$. So I guess, imagine that we're talking about the difference of two Vector Fields that comes out to be zero.

If all of the components of V(x) are zero, then all of the components of V(y) are zero. With transformations like this, (or more complicated ones) then all of the components of V are zero in the y-coordinates. If you know they're all zero in one coordinate, and you know they are tensors, then you know they will be zero in all coordinates.

Not all objects are tensors. We need to have a concept of a derivative of a tensor. Why is it problematic?

To be continued.

 

[JDoolin]

http://www.youtube.com/watch?v=WtPtxz3ef8U" notes, continued (t=10 minutes to t~25 minutes)

Scalar fields:

We know that a scalar field should be the independent of the coordinate system. i.e.

$$\Phi(y) = \Phi(x)$$
(expanded)
$$\Phi(r,\theta,\varphi) = \Phi(x,y,z)$$
Assuming that x and y correspond to the same point.

We're interested in how the derivative of the scalar field transforms.
$$\frac{\partial \Phi} {dy^n}=\frac{\partial x^n}{\partial y^n} \frac{\partial \Phi}{\partial X^m}$$

(expanded)

$$\begin{pmatrix} \frac{\partial \Phi}{\partial r} \\ \frac{\partial \Phi}{\partial \theta} \\ \frac{\partial \Phi}{\partial \varphi} \\ \end{pmatrix} =\begin{pmatrix} \frac{\partial x}{\partial r} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial r} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial r} \frac{\partial \Phi}{\partial_z} \\ \frac{\partial x}{\partial \theta} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial \theta} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial \theta} \frac{\partial \Phi}{\partial_z} \\ \frac{\partial x}{\partial \varphi} \frac{\partial \Phi}{\partial_x} + \frac{\partial y}{\partial \varphi} \frac{\partial \Phi}{\partial_y}+\frac{\partial z}{\partial \varphi} \frac{\partial \Phi}{\partial_z} \end{pmatrix}$$

Whatever coordinates I use, if the temperature is constant in this room, then it will be constant in every set of coordinates. So if the derivative of temp in thermal equilibrium is zero in cartesian coordinates, it will also be zero in polar, spherical, or shmoolivitz coordinates.

Let's take a vector field, though, that is constant in space. It has the same direction everywheres and has the same length everywheres. (The obvious meaning of saying that a vector field is constant) Now, what can you say about the components of this vector as you move from point to point in this same coordinates, x^1 and x^2. For example, let's take the covariant components (the projections onto the axes.) It looks like what we are saying is that the covariant components of the vector are zero.

Now let's take exactly the same situation, but take it in curvilinear coordinates. We have a constant vector field, but the covariant components (projection onto the axes) are not constant. Because the axes are moving and bending around. The same vector; exactly the same vector, pointing in exactly the same direction, but the components will be different.

Incidentally, if we changed from one cartesian coordinate system to another, the components will remain constant. But if the coordinates are flopping around as you move around then the components will also change.

So $$\frac {\partial V_m(x)}{\partial x^n} = 0 \xrightarrow[]{does not}\frac {\partial V_m(y)}{\partial y^n}=0$$

21:00
The derivatives of scalars make perfectly good vectors, but the derivatives of the components of vectors do not transform as the components of tensors. You have to do something a little bit trickier to define derivatives of tensors and make them tensors. The process is called covariant differentiation. (The covariant is not because it has to do with a lower index, it is a completely different use of the term covariant.) What it means is that it's a form of differentiation you can apply to tensors and get back tensors; you can apply to vectors and get back vectors.

Let's see if we can figure out what kind of thing a covariant derivative might be. Let's assume something is a tensor and then prove that it's not.

22:45
Supposing there is a tensor in x-coordinates (rectangular cartesian coordinates) that is equal to the derivative of some vector with respect to x in the x-coordinate system.
What must it be in the y-coordinate system? Does it or does it not become the derivative

$$T_{mn}(x)=\frac{\partial V_m(x)}{\partial x}\overset{?}{\rightarrow}\frac{\partial V_m(y)}{\partial x}$$

The answer is no.

What must the components be in the y-coordinate system? If it really is a tensor, then the components must be:

$$T_{mn}(y)=\frac{\partial x^r}{\partial y^m}\frac{\partial x^s}{\partial y^n}T_{rs}(x)$$

(expanded with two dimensions, that's)
$$\begin{pmatrix} T_{rr}&T_{r \theta} \\ T_{\theta r} & T_{\theta \theta} \end{pmatrix}= \begin{pmatrix} (\frac{\partial x}{\partial r}\frac{\partial x}{\partial r}T_{xx}+ \frac{\partial x}{\partial r}\frac{\partial y}{\partial r}T_{xy}+ \frac{\partial y}{\partial r}\frac{\partial x}{\partial r}T_{yx}+ \frac{\partial y}{\partial r}\frac{\partial y}{\partial r}T_{yy}) & (\frac{\partial x}{\partial r}\frac{\partial x}{\partial \theta}T_{xx}+ \frac{\partial x}{\partial r}\frac{\partial y}{\partial \theta}T_{xy}+ \frac{\partial y}{\partial r}\frac{\partial x}{\partial \theta}T_{yx}+ \frac{\partial y}{\partial r}\frac{\partial y}{\partial \theta}T_{yy}) \\ (\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial r}T_{xx}+ \frac{\partial x}{\partial \theta}\frac{\partial y}{\partial r}T_{xy}+ \frac{\partial y}{\partial \theta}\frac{\partial x}{\partial r}T_{yx}+ \frac{\partial y}{\partial \theta}\frac{\partial y}{\partial r}T_{yy}) & (\frac{\partial x}{\partial \theta}\frac{\partial x}{\partial \theta}T_{xx}+ \frac{\partial x}{\partial \theta}\frac{\partial y}{\partial \theta}T_{xy}+ \frac{\partial y}{\partial \theta}\frac{\partial x}{\partial \theta}T_{yx}+ \frac{\partial y}{\partial \theta}\frac{\partial y}{\partial \theta}T_{yy}) \end{pmatrix}$$

(I think in three dimensions it would be a 3x3 matrix with nine terms in each one. Getting pretty ugly.)

...to be continued

 

[member 11137]

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

Sorry for intervening in your discussion. But I get myself a fundamental difficulty with that apparently trivial assertion... Since the positioning of the 4-tuple (what we call in fact the geometry) is changing within a GTR approach... is there really a difference between the concept of coordinate system and the concept of event ? I don't know if I am clear enougth. With other words: In a "changing geometry as GTR makes it possible" is not the set of all coordinates an event itself? And is not the deformation of this set also another kind of event?

It was just to trying to develop what JDoolin was perhaps trying to say (?)

 

[JDoolin]

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

Sorry for intervening in your discussion. But I get myself a fundamental difficulty with that apparently trivial assertion... Since the positioning of the 4-tuple (what we call in fact the geometry) is changing within a GTR approach... is there really a difference between the concept of coordinate system and the concept of event ? I don't know if I am clear enough. With other words: In a "changing geometry as GTR makes it possible" is not the set of all coordinates an event itself? And is not the deformation of this set also another kind of event?

It was just to trying to develop what JDoolin was perhaps trying to say (?)

Thanks for the question. I will try to develop what I was trying to say, too:

We have the set of events, represented in a cartesian coordinate system + time. The events are not changing, but their positioning might be changing relative to an observer. For instance, two observers will see this same set of events differently. For instance if they are at different places, then the origin will be different.

The coordinates of the events can also appear to be changing relative to a single observer. An observer is constantly moving forward in time. So from the point of view of the observer, all the events of the universe are moving from the present to his past. (The events slide back into history.)

Another way the events can "move" relative to an observer, is if that observer turns left or right, all of the events in the universe move to the left or to the right (relative to that observer), no matter how far away they are. So in a way, the set is "deformed" as the coordinates (y forward, z up, x right) change relative to the observer.

The next question, is, though, if an observer accelerates, shouldn't all of the events be Lorentz Transformed, no matter how far away they are? I think, absolutely they should. However, widely held opinion is that in General Relativity, space-time can only be considered flat in a "local region" and only in this local region does Special Relativity apply.

I suspect that this is an over-interpretation of the math; that in fact, over the long range average, Special Relativity applies over the whole universe; the far-away space-time deforms in just the same way (under the same transformation) as the local space-time.

P.S. Part of the problem I think I see with the GTR approach is that the coordinates are not changing. When one says that the spacetime is flat, only locally, if I understand correctly, this locality is not only in space, but in time. This may be helpful for determining a geodesic of a particle falling through that space, at that time, but once you limit yourself to a locl space-time, you really aren't talking about much more than a single event. If I'm talking about distant stars I'm currently seeing, something that happened a long time ago and a long way away, I can't limit myself to a local spacetime. I need a global cartesian coordinate system. And I need to apply Special Relativity on this global scale to determine effects like stellar aberration.

 

[member 11137]

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event. QUOTE]

Ok. I know and accept this way of thinking. But do you agree that it only holds true if the set of events and the set of coordinates are different sets? It generally works at a macroscopic scale. It works as long as we are able to say "The event (the particle) moves in that direction relatively to that frame (coordinate system)". But what tell us that, at the pico, fermi,...scale, the events (= the particles) are not part of the background it self and deforming it (background = geometric structure described by a set of coordinates)? I suppose that I probably make a "think" mistake but I do not really understand why and where.

 

[JDoolin]

http://www.youtube.com/watch?v=WtPtxz3ef8U"
Notes from 24:00 to about 50:00

24:00 We have an object which I tell you is a tensor and it happens that in the x-coordinate system, its components are the derivatives of V with respect to X_n.

$$T_{mn}(x)=\frac{\partial V_m(x)}{\partial x^n}$$


Ask what its components must be in the y-system. If it really is a tensor, then its componets must be:

$$T_{mn}(y)=\frac{\partial x^r}{\partial y^m} \frac{\partial x^s}{\partial y^n} T_{rs}(x)$$


That is under the assumption that T_mn is a tensor. in the y-coordinates, it must be:

If you know what it is in the x-coordinates, you know automatically what it is in the x coordinates. Now we can substitute the assumed value of T_rs(x):

$$T_{mn}(y)=\frac{\partial x^r}{\partial y^m} \frac{\partial x^s}{\partial y^n} \frac{\partial V_r(x)}{\partial x^s}$$


And my question is this: Is this, or is this not, equal to the derivative of V_m(y) with respect to y_n

$$T_{mn}(y)=\frac{\partial x^r}{\partial y^m} \frac{\partial x^s}{\partial y^n} \frac{\partial V_r(x)}{\partial x^s} ?= \frac{\partial V_m(y)}{\partial y^m}$$


The equals with the question mark doesn't say it is equal, it says, "is it equal?" Is this true? This is how we might have thought, just as derivatives of scalars yield vectors, we might have thought that derivatives of components of vectors transform as tensors, so we're checking out, is this object over here equal to this.

27:00 So let's see if we can work it out. It's not terribly hard. What can we do with this?

$$\frac{\partial x^s}{\partial y^n} \left (\frac{\partial x^s}{\partial y^n} \frac{\partial V_r(x)}{\partial x^s} \right )$$

The part in parentheses, does that ring a bell?

That's the chain rule, so

$$\frac{\partial x^r}{\partial y^m} \frac{\partial V_r(x)}{\partial y^n}$$


I'm asking are those the same? No no no no no. Is that what I'm asking? Yeah, I guess that's what I'm asking.



29:00 Over here I have something involving Vy. Let's change everything on the right-hand-side so that
$$\frac{\partial V_m(y)}{\partial y^m}= \frac{\partial}{\partial y^n} \frac{\partial x^r}{\partial y_m} V_r(x)$$


All I've done here is was use the transformation from y-components to x-components to rewrite Vm() as a derivative of Vm().

$$\frac{\partial}{\partial y^n} \left(\frac{\partial x^r}{\partial y_m}V_r(x) \right) ?=\frac{\partial x^r}{\partial y^m} \frac{\partial V_r(x)}{\partial y^n}$$


Okay, well, let's look at it.
We have here a derivative of a product. So first of all, the derivative will hit V. That will give me a term that looks exactly like the right-hand-side. But its not good, because it's only one of the terms when I take the derivative here. (Product Rule)


$$\frac{\partial}{\partial y^n} \left(\frac{\partial x^r}{\partial y_m}V_r(x) \right) = \frac{\partial}{\partial y^n } \left (\frac{\partial}{ \partial y^m}\right ) V_r(x) + \frac{\partial x^r}{\partial y^m} \frac{\partial V_r(x)}{\partial y^n}$$


33:00 the answer is, it is not equal to the right hand side. In other words, this thing that I have called a tensor has not transformed as a tensor. If I insist that T_mn(x) are the x-components of a tensor then It is not true that the y-components of the tensor are just derivatives of the tensor with respect to y.

Not true.

Let's see what we can say. (Answering Question: Just because the components of a vector may be constant doesn't mean it is the same in alll "frames of references" (coordinate systems?) If the x-coordinates are linear in the y-coordinates, then the term would be zero. It is because the relationship between the coordinates are varying from place to place. If they vary, then you get that extra term, and it's proportional to V with no derivative.

If T really transforms as a tensor, then you don't get the extra term. That suggests that we have to do something just a little bit different.

37:00 The right hand side is by definition T_mn(y). We forced it to transform in the right way. What we see is that it has this extra term over here. It's not just what you might have thought it was. Let's try to guess then, what the right way to differentiate tensors is. The right way must have an extra term to cancel this out. Let's give the extra term a name. capital gamma

$$\Gamma_{nm}^r$$


Here's what we have discovered. That if we want to differentiate a tensor we have to do something just a little bit trickier than just take the derivative of its components.

$$\frac{\partial V_m}{\partial y^n} +\Gamma_{nm}^r V_r$$


You have to take into account the fact that the coordinates are varying from point to point. Let's suppose (we're making a guess) that if we could find the right object gamma, we could do a new kind of derivative; a new kind of tensor which we could call T_mn. this process of differentiating is called covariant derivative. It is written with an upside down triangle (nabla)

$$T_{mn} = \nabla_n V_m = \frac{\partial V_m}{\partial y^n} +\Gamma_{nm}^r V_r$$


41:00 It's going to depend not only on the metric tensor, but on the derivatives of the metric tensor. If the metric tensor has derivatives (if the metric tensor is not constant) it means in some way the coordinates are flopping around. Is there an object which has the property that when we construct the covariant derivative what occurs is actually a tensor. that it really acts as a tensor and transforms as a tensor?

That will be a brilliant achievement, because then we'll be able to write equations such as the covariant derivative of a vector might be equal to some other tensor, maybe a completely different object

The lHS could be the derivative of an electric field. The RHS could be some other tensor of some other physical character

we might have a situation The covariant derivative of the electric field... is equal to some tensor

Will it be true in all reference frames? Yes, if the left-hand-side is a tensor and transforms as a tensor.

Can we find this Gamma

Will it be a tensor in all reference frames? yes if the LHS operates as a tensor and transforms as a tensor.l

It can happen that we might find a system that can't be described in cartesian coordinate system.

I'll give you the general rule (for things with lower components) Because the coordinates vary from place to place, there are contributions of this type for each one of the components.

At some point, I'm going to leave for you a bunch of algebra to work out.But I'll give you the rules.

What would happen if I wanted to make a covariant derivative of a tensor. Incidentally, the derivative of a scalar is just an ordinary derivative.

Let's take a tensor with two indexes, and supposing we want to take its covariant derivative with respect to y. I'll tell you what I want to change notation a little bit here.

$$T_{mp} = \nabla_p V_m = \frac{\partial V_m}{\partial y^p} +\Gamma_{pm}^r V_r$$


Same thing twice with extra index along for the ride.

$$\nabla_p T_{mn}=\frac{\partial T_mn}{\partial y^p}+ \Gamma_{pm}^r T_{rn}+ \Gamma_{pn}^r T_{mr}$$


For each index in the tensor, No matter how many of them there are, you get another term with the gamma, using just one of the indices, turning a blind eye to all the other indices.

 

[JDoolin]

A coordinate system is just a function that assigns a 4-tuple of real numbers to each event.

Ok. I know and accept this way of thinking. But do you agree that it only holds true if the set of events and the set of coordinates are different sets? It generally works at a macroscopic scale. It works as long as we are able to say "The event (the particle) moves in that direction relatively to that frame (coordinate system)". But what tell us that, at the pico, fermi,...scale, the events (= the particles) are not part of the background it self and deforming it (background = geometric structure described by a set of coordinates)? I suppose that I probably make a "think" mistake but I do not really understand why and where.

When you say "The event (the particle)" this confuses me. An event is not a particle. Events are things that happen to particles at a certain place and a certain time.

Now, as for how and whether particles deform the geometric structure of spacetime, that's the question I'm pursuing. I have no argument, for instance, against the idea that particles follow geodesics; but if one were to claim that an orbital geodesic is actually a straight line, I would be concerned.

 

[pervect]

I have no such desire to free myself from this idea. To me, any legitimate coordinate system should have immediate metrical meaning. And all coordinate transformations simply change the metrical meanings.

What's an "immediate metrical meaning"? If you're on the surface of the Earth, do lattitude and longitude have an "immediate metrical meaning"? I find that thinking about the issues that arise because the Earth has a curved surface is a good way to think about the same issues that arise in space-time.

When I think of the distance between two points, I think of the shortest distance between two points. Integrating along the path to find the distance between two points represents the distance traveled along the path between the two points. I don't see how this is a change in space-time. It is a change in what you mean by distance.

Well, imagine that you're on the surface of the Earth again, (it shouldn't be hard , cause you are :-) ), and you want the distance between two cities. Do you insist that the distance be measured by a straight line path, that bores underground, or can you be comfortable with measuring the distance "on the surface"? The shortest distance on the surface will turn out to be measured along a great circle, which is the shortest path connecting two points on the Earth that lies on the surface.


If someone told you that city A was such and such a distance from city B - would you expect that they were talking about the distance through a tunnel drilled between them, or would you think that they were talking about the distance on the surface?

 

[JDoolin]

What's an "immediate metrical meaning"? If you're on the surface of the Earth, do lattitude and longitude have an "immediate metrical meaning"? I find that thinking about the issues that arise because the Earth has a curved surface is a good way to think about the same issues that arise in space-time.



Well, imagine that you're on the surface of the Earth again, (it shouldn't be hard , cause you are :-) ), and you want the distance between two cities. Do you insist that the distance be measured by a straight line path, that bores underground, or can you be comfortable with measuring the distance "on the surface"? The shortest distance on the surface will turn out to be measured along a great circle, which is the shortest path connecting two points on the Earth that lies on the surface.


If someone told you that city A was such and such a distance from city B - would you expect that they were talking about the distance through a tunnel drilled between them, or would you think that they were talking about the distance on the surface?

Certainly, if someone told me the distance between two cities, I would expect them to be talking about the distance along a particular road or set of roads. But where are you taking this? The real distance between the two points is the one connecting the two cities right through the surface of the planet.

Yes, the latitude and longitude have real metrical meaning, and one knows that an arc-segment of a certain angle on a circle has a length of the radius times that angle in radians. But you don't throw away the knowledge of the length of the secant line between the two points. And there's not really any question of which one is the real distance.

We don't (or we shouldn't, anyway) claim that the space is curved. Certainly the road is curved, and the surface of the Earth is curved, but where is it curved? It is curved in plain old cartesian coordinates. We live in a plain old cartesian coordinate system universe. Though we can show that objects fall toward planets, and even light bends around or into stars, it's within a cartesian coordinate system that all of this curving and bending can be observed.

 

[pervect]

There isn't anything particularly "unreal" about the distance along the surface from my point of view. It's just as "real" as the other sort, it's just defined differently. Presumably, though, you've got some particular mental model or framework that you're comfortable with, and that's the actual issue. You call things that fit neatly together with your own personal model "real". People can (and do) argue about what's "real" endlessly because everyone has their own internal mental models, and if you do so, it's just a total distraction, and one never gets anywhere.

If you can learn to work with the abstractions of lattitude and longitude, though, there's no reason why you can't learn how to deal with generalized coordinates. Even if they're not "real", you can do the math. And there's a reason to do the math, because it's useful to work with what you're interested in (distances along the surface of the Earth) rather than things you're not interested in (distances that you could travel along if you had a high speed tunnel borer, and didn't have to worry about the havok that actually drilling the holes would cause).As far as whether space or not is curved - while the answer is that GR predicts yes (depending actually on how you slice it), the more important question is really whether or not space-time is curved.

So, it might be helpful to think about how we determine if something is curved operationally or not. A very short answer is that if you go 500 miles east and 300 miles north on the Earth's surface, in that order, you wind up at a different point than when you go 300 miles north and 500 miles east. This is something that doesn't happen in a planar geometry. It happens only when you are on a "curved" geometry.

To see this imagine starting out at a point on the equator. The change in the longitude coordinate will depend on your latitude, so you'll get a smaller change in longitude by going 500 miles east on the equator than you'll get by going 500 miles east at a higher lattitude.

I suppose it's OK to think of a "curved" geometry as "less real" if you absolutely must, as long as you're able to work with it and understand the examples and tell when a geometry can be a "real, good-old fashioned Euclidean one" as opposed to the "made up " one that "isn't really truly real". But it's both polite and more communicative to say to other people "Non-euclidean geometry" rather than "made up unreal geometry" - all the mathemeticians who study it will tell you (with a good deal of justice, I'd add) that their geometry is just as real as yours is.

Anyway - I digress. The point is that if you go 500 miles east, and 300 miles north, you wind up at a different point on the Earth's surface. There's an analogous situation in space-time.

If you go 500 seconds into the future, and then 300 miles up, in that order, you wind up at a different point in space-time than you do when you arrive at when you go 300 miles up first and then go 500 seconds into the future. That's because of gravitational time dilation - the clocks run at different rates at higher elevations.

So, we deal with this by saying that the geometry of space-time is curved. And we introduce a metric coefficient to turn coordinates into distances, just as we do on the surface of the Earth.

There should be in theory similar effects (where the order of motion matters) purely in space - but they're much smaller in magnitude (and more subtly, it depends on the spatial slice - how you define simultaneity). The more obvious, "can't avoid it" issue is in the behavior of clocks, however.

 

[JDoolin]

There isn't anything particularly "unreal" about the distance along the surface from my point of view. It's just as "real" as the other sort, it's just defined differently. Presumably, though, you've got some particular mental model or framework that you're comfortable with, and that's the actual issue. You call things that fit neatly together with your own personal model "real". People can (and do) argue about what's "real" endlessly because everyone has their own internal mental models, and if you do so, it's just a total distraction, and one never gets anywhere.

If you can learn to work with the abstractions of lattitude and longitude, though, there's no reason why you can't learn how to deal with generalized coordinates. Even if they're not "real", you can do the math. And there's a reason to do the math, because it's useful to work with what you're interested in (distances along the surface of the Earth) rather than things you're not interested in (distances that you could travel along if you had a high speed tunnel borer, and didn't have to worry about the havok that actually drilling the holes would cause).

As far as whether space or not is curved - while the answer is that GR predicts yes (depending actually on how you slice it), the more important question is really whether or not space-time is curved.

So, it might be helpful to think about how we determine if something is curved operationally or not. A very short answer is that if you go 500 miles east and 300 miles north on the Earth's surface, in that order, you wind up at a different point than when you go 300 miles north and 500 miles east. This is something that doesn't happen in a planar geometry. It happens only when you are on a "curved" geometry.

To see this imagine starting out at a point on the equator. The change in the longitude coordinate will depend on your latitude, so you'll get a smaller change in longitude by going 500 miles east on the equator than you'll get by going 500 miles east at a higher lattitude.

I suppose it's OK to think of a "curved" geometry as "less real" if you absolutely must, as long as you're able to work with it and understand the examples and tell when a geometry can be a "real, good-old fashioned Euclidean one" as opposed to the "made up " one that "isn't really truly real". But it's both polite and more communicative to say to other people "Non-euclidean geometry" rather than "made up unreal geometry" - all the mathemeticians who study it will tell you (with a good deal of justice, I'd add) that their geometry is just as real as yours is.

Anyway - I digress. The point is that if you go 500 miles east, and 300 miles north, you wind up at a different point on the Earth's surface. There's an analogous situation in space-time.

If you go 500 seconds into the future, and then 300 miles up, in that order, you wind up at a different point in space-time than you do when you arrive at when you go 300 miles up first and then go 500 seconds into the future. That's because of gravitational time dilation - the clocks run at different rates at higher elevations.

So, we deal with this by saying that the geometry of space-time is curved. And we introduce a metric coefficient to turn coordinates into distances, just as we do on the surface of the Earth.

There should be in theory similar effects (where the order of motion matters) purely in space - but they're much smaller in magnitude (and more subtly, it depends on the spatial slice - how you define simultaneity). The more obvious, "can't avoid it" issue is in the behavior of clocks, however.

I'm sorry about going on and on about "real." Perhaps it is just a distraction. But I have to defend the use of the words "actually" and "really" because I believe that physics describes reality. I believe that we live in a real universe that has real properties. If you want to define distance to mean the distance traveled by a particle between objects, and I want to define distance to mean the straight-line distance between the objects. You are right in saying that one definition is just as "real" as the other. One defines the straight-line distance, and the other defines the traveled-distance.

But another question would be to ask "what direction is the next city" One person (comfortable with driving) points along the road that curves several times before getting to the city. Another person (comfortable with non-Euclidian geometry)-points straight, tangentially along the curve of the Earth to a point on the horizon that passes several thousand feet above the city. A third person (comfortable with Euclidian geometry) points straight, through the planet toward the city. Which person is correct? All three people are comfortable in their particular mental framework. But it can still be asked "which way are they really pointing" The first is actually pointing along a road. The second is actually pointing at a distant star. The third is actually pointing toward the city.

In the Leonard Susskind GR lecture 5, about 18:30, someone asks about the existence of a "third coordinate system." Susskind responds by talking about good-old cartesian coordinates vs. curvilinear coordinate systems. If the coordinates are moving around and flopping around then vectors which are "actually" constant end up looking non-constant. If we have good-old cartesian coordinates, constant vectors look constant.

Dr. Susskind uses the word "actually." He doesn't mean reality should be compared to some mysterious third "actual" coordinate system. He is making a circular relationship (one that I think is quite appropriate). "actually constant" means "constant in cartesian coordinates" and "constant in cartesian coordinates" means "actually constant."

I don't think that anyone would go in the other direction. For instance, if I said that a plane going around the Earth "looks like" it is going around the equator, but it is "actually" a straight line in spherical coordinates. This doesn't make sense. If you're using spherical coordinates, you know that your great circles are not "really" straight lines. You call them great circles because that's what they actually are; circles. It's called a non-Euclidean geometry, because the "parallel lines" cross, but you know that what you're calling parallel lines aren't "really" lines, and they aren't "really" parallel.

If you think that the problem here is with the word "really," that misses the more important that words have actual meanings, and to talk about non-Euclidean geometry, you're deliberately changing the meanings of those words. "all the mathemeticians who study it will tell you (with a good deal of justice, I'd add) that their geometry is just as real as yours is." Is a line around the surface of a sphere really a line? Is it really straight? Are the longitude lines around the Earth really parallel? Will the mathematicians who study non-Euclidian geometry answer all of these questions with an emphatic "yes?"

Even if the geometry, itself, is real, I think that most will acknowledge that a real straight line is one that is straight in cartesian space, and the great circles on surfaces of spheres are really not straight.

 

[Fredrik]

Now, as for how and whether particles deform the geometric structure of spacetime, that's the question I'm pursuing. I have no argument, for instance, against the idea that particles follow geodesics; but if one were to claim that an orbital geodesic is actually a straight line, I would be concerned.

A straight line in $\mathbb R^2$ can be defined as a curve with zero acceleration. $x:[a,b]\rightarrow\mathbb R^2$ and x''(t)=0 for all t in [a,b]. A "straight line" in a manifold can't be defined exactly like that, since there's no addition operation defined on a manifold. But the velocity vector field of a curve and the covariant derivative along a curve can both be defined in a coordinate-independent way. So we can define a "straight line" in a manifold to be a curve such that the covariant derivative along the curve, of the curve's velocity vector field along the curve, is =0. These curves are what correspond to straight lines in $\mathbb R^2$, but we don't call them "straight lines", we call them "geodesics".

The best place to read about this is "Riemannian manifolds: an introduction to curvature", by John M. Lee.

The world line of a planet in orbit around a star is a geodesic, and if someone wants to call a geodesic a "straight line", I wouldn't have a problem with that.

 

[PeterDonis]

Even if the geometry, itself, is real, I think that most will acknowledge that a real straight line is one that is straight in cartesian space, and the great circles on surfaces of spheres are really not straight.

The reason we can make this distinction, though, is that we know the Earth's surface is a 2-dimensional manifold that is embedded in a 3-dimensional space that is (as far as we can tell) Euclidean (or what you're calling "cartesian", though that term really applies to a specific set of coordinates, not to the underlying geometric space). So it seems natural to us to identify the straight lines in the 3-dimensional space as "straight", but not the lines that are geodesics in the 2-dimensional manifold.

However, there is no guarantee that we will always be able to find a higher-dimensional Euclidean space in which a given manifold is embedded. For example, it is possible (though not likely, given the current state of evidence) that the 3-dimensional manifold that we call "space" is actually not Euclidean--that it has some (very small) intrinsic curvature, so that what you are calling "straight lines" in that manifold, like the line that goes through the interior of the Earth to get from, say, New York to London, will turn out *not* to be "straight lines" when we are able to make accurate enough measurements; instead, they will turn out to be (small segments of) geodesics in a curved manifold. Furthermore, it is possible that there is *no* higher-dimensional Euclidean manifold in which the 3-dimensional space of our experience is embedded. (String theory says that the full underlying manifold of the universe has many more dimensions, but as far as I know, the full underlying manifold is not Euclidean either.) Certainly there is no logical requirement that any non-Euclidean manifold *has* to be embedded in a higher-dimensional Euclidean one.

So the distinction you're making, that only straight lines in a Euclidean manifold qualify as "real" straight lines, is really dependent on the contingent fact that the 3-dimensional space of our experience happens to be Euclidean, at least to the accuracy of our current measurements. To put it another way, I have no problem with adopting a terminology that only applies the term "straight line" to straight lines in a Euclidean manifold, as long as it is understood that that terminology is just a convenience and does not reflect any greater "reality" attributed to Euclidean straight lines as opposed to geodesics in other manifolds.